伯努利数

伯努利数是一个用于解决 nn 次方和的数列。

它的递归定义公式如下:

i=0n(n+1i)Bi=[n=0]        (1.1)\sum_{i=0}^n \binom {n+1} i B_i=[n=0] ~~~~~~~~ (1.1)

通过这个定义可以得到伯努利数的前几项:1,12,16,0...1,-\frac{1}{2},\frac{1}{6},0...

Sm(n)=i=0n1imS_m(n)=\sum_{i=0}^{n-1} i^m ,伯努利通过找规律发现了伯努利公式:

Sm(n)=1m+1i=0m(m+1i)Binm+1i        (1.2)S_m(n)=\frac{1}{m+1} \sum_{i=0}^m \binom {m+1} i B_i n^{m+1-i} ~~~~~~~~ (1.2)

《具体数学》上给出的证明如下:

Sm+1(n)+nm+1=i=0nim+1=i=0n1(i+1)m+1S_{m+1}(n)+n^{m+1}=\sum_{i=0}^ni^{m+1}=\sum_{i=0}^{n-1} (i+1)^{m+1}

=i=0n1j=0m+1(m+1j)ij=\sum_{i=0}^{n-1}\sum_{j=0}^{m+1} \binom {m+1}{j} i^j

=j=0m+1(m+1j)i=0n1ij=\sum_{j=0}^{m+1}\binom {m+1}{j} \sum_{i=0}^{n-1} i^j

                =j=0m+1(m+1j)Sj(n)        (1.3)~~~~~~~~~~~~~~~~ =\sum_{j=0}^{m+1}\binom {m+1}{j} S_j(n) ~~~~~~~~ (1.3)

(1.3)(1.3)两边同时减去得,

Sm+1(n)=j=0m(m+1j)Sj(n)        (1.4)S_{m+1}(n)=\sum_{j=0}^{m}\binom {m+1}{j} S_j(n) ~~~~~~~~ (1.4)

Sm(n)S'_ m(n)(1.2)(1.2) 右式 , Δ=Sm(n)Sm(n)\Delta=S_m(n)-S'_m(n)

归纳证明 Sm(n)=Sm(n)S_m(n)=S'_m(n)

1.当 m=0m=0 时成立。

2.设对于 i[0,m),Si(n)=Si(n)\forall i\in[0,m),S_i(n)=S'_ i(n), 由 (3.4)(3.4) 得:

nm+1=j=0m(m+1j)Sj(n)+(m+1m)Δ      (只有j=m时有差异)n^{m+1}=\sum_{j=0}^{m} \binom{m+1}{j} S'_ {j}(n)+ \binom{m+1}{m} \Delta ~~~~~~ \text{(只有j=m时有差异)}

              =j=0m(m+1j)1j+1k=0j(j+1k)Bknj+1k+(m+1)Δ      (化简代入)~~~~~~~~~~~~~~ =\sum_{j=0}^{m} \binom{m+1}{j} \frac{1}{j+1} \sum_{k=0}^j \binom {j+1} k B_k n^{j+1-k} + (m+1) \Delta ~~~~~~ \text{(化简代入)}

=j=0mk=0j(m+1j)(j+1k)Bkj+1nj+1k+(m+1)Δ=\sum_{j=0}^{m} \sum_{k=0}^j \binom{m+1}{j} \binom {j+1} k \frac{B_k}{j+1} n^{j+1-k} + (m+1) \Delta

              =j=0mk=0j(m+1j)(j+1jk)Bjkj+1nk+1+(m+1)Δ      (将k换为j-k)~~~~~~~~~~~~~~ =\sum_{j=0}^{m} \sum_{k=0}^j \binom{m+1}{j} \binom {j+1} {j-k} \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta ~~~~~~ \text{(将k换为j-k)}

              =j=0mk=0j(m+1j)(j+1k+1)Bjkj+1nk+1+(m+1)Δ      ((nm)=(nnm))~~~~~~~~~~~~~~ =\sum_{j=0}^{m} \sum_{k=0}^j \binom{m+1}{j} \binom {j+1} {k+1} \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta ~~~~~~ (\binom {n}{m} = \binom {n}{n-m})

=k=0mj=km(m+1j)(j+1k+1)Bjkj+1nk+1+(m+1)Δ=\sum_{k=0}^{m} \sum_{j=k}^{m} \binom{m+1}{j} \binom {j+1} {k+1} \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta

               =k=0mj=km(m+1j)(jk)j+1k+1Bjkj+1nk+1+(m+1)Δ      ((n+1m+1)=(nm)×n+1m+1)~~~~~~~~~~~~~~~ =\sum_{k=0}^{m} \sum_{j=k}^{m} \binom{m+1}{j} \binom {j} {k} * \frac{j+1}{k+1} * \frac{B_{j-k}}{j+1} n^{k+1} + (m+1) \Delta ~~~~~~ (\binom{n+1}{m+1}=\binom{n}{m} \times \frac{n+1}{m+1})

=k=0mnk+1k+1j=km(m+1j)(jk)Bjk+(m+1)Δ=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \sum_{j=k}^{m} \binom{m+1}{j} \binom {j} {k} B_{j-k} + (m+1) \Delta

=k=0mnk+1k+1(m+1k)j=km(m+1kjk)Bjk+(m+1)Δ=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \binom{m+1}{k} \sum_{j=k}^{m} \binom{m+1-k}{j-k} B_{j-k} + (m+1) \Delta

=k=0mnk+1k+1(m+1k)j=0mk(m+1kj)Bj+(m+1)Δ=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \binom{m+1}{k} \sum_{j=0}^{m-k} \binom{m+1-k}{j} B_{j} + (m+1) \Delta

=k=0mnk+1k+1(m+1k)[mk=0]+(m+1)Δ     (将(1.1)=\sum_{k=0}^{m} \frac{n^{k+1}}{k+1} \binom{m+1}{k} [m-k=0] + (m+1) \Delta ~~~~~ (\text{将(1.1)}带

=nm+1m+1(m+1m)+(m+1)Δ=\frac{n^{m+1}}{m+1} \binom{m+1}{m} + (m+1) \Delta

=nm+1+(m+1)Δ=n^{m+1} + (m+1) \Delta

所以 (m+1)Δ=0(m+1)\Delta=0 , 又 m1m \ge 1 , 所以 Δ=0\Delta=0 , 证毕。